Derivatives

From average rate to instantaneous rate. The average rate of change of $f$ between $x$ and $x+h$ is the familiar slope-of-a-chord expression $\frac{f(x+h)-f(x)}{h}$. This is exact but describes an interval, not a single point. Shrinking $h$ toward $0$ turns that chord into a tangent line touching the curve at exactly one point, and the limiting value of the slope is the derivative:

where: $f'(x)$ is the derivative of $f$ at $x$ (also written $\frac{dy}{dx}$ or $\frac{d}{dx}f(x)$), $h$ is a small increment in $x$ that is never actually set to zero but only driven toward it, and the limit must exist for $f$ to be called differentiable at $x$. Computing $f'$ this way is called differentiating from first principles or ab initio.

Two readings of the same number. Geometrically, $f'(x)$ is the slope of the tangent to $y=f(x)$ at that point — positive where the curve rises, negative where it falls, zero at a flat point. Physically, if $f(t)$ is the position of a particle at time $t$, $f'(t)$ is its instantaneous velocity — the speed a speedometer would show at that exact instant, not averaged over a stretch of time. Both readings describe the same limiting number; only the meaning of $x$ (a position on a graph, or a moment in time) changes with the context.

Differentiability is not automatic. A function can fail to have a derivative at a point even where it is perfectly well defined there — the limit in the definition simply may not exist. This happens, for instance, at a sharp corner on a graph, where approaching from the left and from the right gives two different limiting slopes, so no single tangent line is well defined. At this level it is enough to know that "differentiable at $x$" is a real condition worth checking, not something every function automatically satisfies; a full treatment of continuity and differentiability, with left-hand and right-hand derivatives, belongs to Class 12.

Algebra of derivatives. If $u(x)$ and $v(x)$ are differentiable, the derivative distributes over sums and differences, and follows fixed patterns for products and quotients:

where: $u'(x)$ and $v'(x)$ are the separate derivatives of $u$ and $v$; note the product rule is not $u'v'$, and the quotient rule's numerator order matters since subtraction is not commutative.

Standard derivatives. At this level only polynomial and trigonometric functions are built from first principles and then reused as building blocks:

where: $n$ is any real exponent and $c$ is a fixed constant. These four results themselves come from the first-principles limit combined with two standard limits proved earlier in the chapter, $\lim_{x\to a}\frac{x^n-a^n}{x-a}=na^{n-1}$ and $\lim_{x\to0}\frac{\sin x}{x}=1$ — so nothing here is assumed without justification, only compressed for speed.

Find $f'(x)$ for $f(x)=x^2+3x$, and use it to find the slope of the tangent to $y=x^2+3x$ at $x=2$; then find the derivative of $g(x)=x\sin x$ using the product rule.

Step 1. From the derivation above, $f'(x)=2x+3$ for every $x$. Substituting $x=2$: $f'(2)=2(2)+3=7$. So the tangent to $y=x^2+3x$ at $x=2$ has slope $\boldsymbol{7}$.

Step 2. For $g(x)=x\sin x$, identify the two factors for the product rule: $u(x)=x$ and $v(x)=\sin x$, with $u'(x)=1$ (the derivative of $x$ itself) and $v'(x)=\cos x$ (a standard result).

Step 3. Apply $\frac{d}{dx}(uv)=u'v+uv'$: $g'(x)=(1)(\sin x)+(x)(\cos x)=\sin x+x\cos x$.

Final answer: $f'(x)=2x+3$ with tangent slope $\boldsymbol{7}$ at $x=2$; and $\boldsymbol{g'(x)=\sin x+x\cos x}$.

Competitive stretch

As a quick check on $g'(x)=\sin x+x\cos x$, evaluate at $x=0$: $g'(0)=\sin0+0\cdot\cos0=0$. This is consistent, since $x\sin x\geq0$ on both sides of $x=0$ near the origin (a negative $x$ makes $\sin x$ negative too, so the product stays non-negative), meaning $y=x\sin x$ behaves like it has a minimum there — a horizontal tangent, matching $g'(0)=0$.

• Stopping at $\frac{f(x+h)-f(x)}{h}$ and calling it "the derivative" — this expression is only the average rate of change until the limit $h\to0$ is actually taken.
• Applying the product rule as $u'v'$ (multiplying the two derivatives together) instead of the correct $u'v+uv'$.
• Getting the quotient rule's numerator order wrong, or forgetting the $[v(x)]^2$ denominator — since subtraction is not commutative, $u'v-uv'$ and $uv'-u'v$ are not the same thing.
• Reaching for the chain rule or trying to differentiate $e^x$, $\ln x$, or inverse trigonometric functions — all of these are Class 12 material; this chapter's scope is polynomial and trigonometric functions built with sum, product and quotient rules only.

Ask the tutor: want another function differentiated from first principles, side by side with the shortcut-rule version, to see exactly where the two methods meet?